I believe you meant to write
[tex]log_{2}(x+2) + log_{2}(x-2) \leq log_{2}(5) [/tex] ?
If that's the case I'll solve the one I provided but I'll drop the base 2 to type it faster but you need to put it always!
Remember: log a + log b = log (a*b)
So log (x+2) + log (x-2) = log [(x+2)*(x-2)] = log (x^2 - 4)
Now back to the inequality:
log (x^2 - 4) ≤ log 5
Raise both sides as powers of 2 ( Since it's the base of your log)
Now,
x^2 - 4 ≤ 5
Add 4 both sides:
x^2 ≤ 9
Square root both sides
x ≤ +3 or x ≤ -3
Reject the -3 solution as it makes both (x + 2) and (x - 2) negative and a log can never have a negative value inside its brackets.
So x ≤ 3 But can never be less than 2 as well for the same previous reason.
Hope that helped.
